Steel Construction: Manual Design-examples

DESIGN EXAMPLES
Version 14.0
AMERICAN INSTITUTE
OF
STEEL CONSTRUCTION

Design Examples V14.0
AMERICAN INSTITUTE OF STEEL CONSTRUCTION
ii
Copyright © 2011
by
American Institute of Steel Construction
All rights reserved. This publication or any part thereof
must not be reproduced in any form without the
written permission of the publisher.
The AISC logo is a registered trademark of AISC.
The information presented in this publication has been prepared in accordance with recognized engineering principles
and is for general information only. While it is believed to be accurate, this information should not be used or relied
upon for any specific application without competent professional examination and verification of its accuracy,
suitability, and applicability by a licensed professional engineer, designer, or architect. The publication of the material
contained herein is not intended as a representation or warranty on the part of the American Institute of Steel
Construction or of any other person named herein, that this information is suitable for any general or particular use or
of freedom from infringement of any patent or patents. Anyone making use of this information assumes all liability
arising from such use.
Caution must be exercised when relying upon other specifications and codes developed by other bodies and
incorporated by reference herein since such material may be modified or amended from time to time subsequent to the
printing of this edition. The Institute bears no responsibility for such material other than to refer to it and incorporate it
by reference at the time of the initial publication of this edition.
Printed in the United States of America
First Printing, October 2011

iii
PREFACE
The primary objective of these design examples is to provide illustrations of the use of the 2010 AISC Specification
for Structural Steel Buildings (ANSI/AISC 360-10) and the 14th Edition of the AISC Steel Construction Manual.
The design examples provide coverage of all applicable limit states whether or not a particular limit state controls the
design of the member or connection.
In addition to the examples which demonstrate the use of the Manual tables, design examples are provided for
connection designs beyond the scope of the tables in the Manual. These design examples are intended to demonstrate
an approach to the design, and are not intended to suggest that the approach presented is the only approach. The
committee responsible for the development of these design examples recognizes that designers have alternate
approaches that work best for them and their projects. Design approaches that differ from those presented in these
examples are considered viable as long as the Specification, sound engineering, and project specific requirements are
satisfied.
Part I of these examples is organized to correspond with the organization of the Specification. The Chapter titles
match the corresponding chapters in the Specification.
Part II is devoted primarily to connection examples that draw on the tables from the Manual, recommended design
procedures, and the breadth of the Specification. The chapters of Part II are labeled II-A, II-B, II-C, etc.
Part III addresses aspects of design that are linked to the performance of a building as a whole. This includes
coverage of lateral stability and second order analysis, illustrated through a four-story braced-frame and moment-
frame building.
The Design Examples are arranged with LRFD and ASD designs presented side by side, for consistency with the
AISC Manual. Design with ASD and LRFD are based on the same nominal strength for each element so that the
only differences between the approaches are which set of load combinations from ASCE/SEI 7-10 are used for
design and whether the resistance factor for LRFD or the safety factor for ASD is used.
CONVENTIONS
The following conventions are used throughout these examples:
1. The 2010 AISC Specification for Structural Steel Buildings is referred to as the AISC Specification and the
14th Edition AISC Steel Construction Manual, is referred to as the AISC Manual.
2. The source of equations or tabulated values taken from the AISC Specification or AISC Manual is noted
along the right-hand edge of the page.
3. When the design process differs between LRFD and ASD, the designs equations are presented side-by-side.
This rarely occurs, except when the resistance factor, φ, and the safety factor, Ω, are applied.
4. The results of design equations are presented to three significant figures throughout these calculations.
ACKNOWLEDGMENTS
The AISC Committee on Manuals reviewed and approved V14.0 of the AISC Design Examples:
William A. Thornton, Chairman
Mark V. Holland, Vice Chairman
Abbas Aminmansour
Charles J. Carter
Harry A. Cole
Douglas B. Davis
Robert O. Disque
Bo Dowswell
Edward M. Egan
Marshall T. Ferrell

iv
Lanny J. Flynn
Patrick J. Fortney
Louis F. Geschwindner
W. Scott Goodrich
Christopher M. Hewitt
W. Steven Hofmeister
Bill R. Lindley, II
Ronald L. Meng
Larry S. Muir
Thomas M. Murray
Charles R. Page
Davis G. Parsons, II
Rafael Sabelli
Clifford W. Schwinger
William N. Scott
William T. Segui
Victor Shneur
Marc L. Sorenson
Gary C. Violette
Michael A. West
Ronald G. Yeager
Cynthia J. Duncan, Secretary
The AISC Committee on Manuals gratefully acknowledges the contributions of the following individuals who
assisted in the development of this document: Leigh Arber, Eric Bolin, Janet Cummins, Thomas Dehlin, William
Jacobs, Richard C. Kaehler, Margaret Matthew, Heath Mitchell, Thomas J. Schlafly, and Sriramulu Vinnakota.

v
TABLE OF CONTENTS
PART I. EXAMPLES BASED ON THE AISC SPECIFICATION
CHAPTER A GENERAL PROVISIONS.........................................................................................................A-1
Chapter A
References ......................................................................................................................................................A-2
CHAPTER B DESIGN REQUIREMENTS .....................................................................................................B-1
Chapter B
References ......................................................................................................................................................B-2
CHAPTER C DESIGN FOR STABILITY.......................................................................................................C-1
Example C.1A Design of a Moment Frame by the Direct Analysis Method ........................................................C-2
Example C.1B Design of a Moment Frame by the Effective Length Method ......................................................C-6
Example C.1C Design of a Moment Frame by the First-Order Method .............................................................C-11
CHAPTER D DESIGN OF MEMBERS FOR TENSION...............................................................................D-1
Example D.1 W-Shape Tension Member ..........................................................................................................D-2
Example D.2 Single Angle Tension Member ....................................................................................................D-5
Example D.3 WT-Shape Tension Member ........................................................................................................D-8
Example D.4 Rectangular HSS Tension Member ............................................................................................D-11
Example D.5 Round HSS Tension Member ....................................................................................................D-14
Example D.6 Double Angle Tension Member .................................................................................................D-17
Example D.7 Pin-Connected Tension Member ...............................................................................................D-20
Example D.8 Eyebar Tension Member ............................................................................................................D-23
Example D.9 Plate with Staggered Bolts .........................................................................................................D-25
CHAPTER E DESIGN OF MEMBERS FOR COMPRESSION...................................................................E-1
Example E.1A W-Shape Column Design with Pinned Ends ............................................................................... E-4
Example E.1B W-Shape Column Design with Intermediate Bracing.................................................................. E-6
Example E.1C W-Shape Available Strength Calculation .................................................................................... E-8
Example E.1D W-Shape Available Strength Calculation .................................................................................... E-9
Example E.2 Built-up Column with a Slender Web........................................................................................ E-11
Example E.3 Built-up Column with Slender Flanges ...................................................................................... E-16
Example E.4A W-Shape Compression Member (Moment Frame) .................................................................... E-21
Example E.4B W-Shape Compression Member (Moment Frame) .................................................................... E-25
Example E.5 Double Angle Compression Member without Slender Elements ............................................... E-26
Example E.6 Double Angle Compression Member with Slender Elements .................................................... E-31
Example E.7 WT Compression Member without Slender Elements ............................................................... E-37
Example E.8 WT Compression Member with Slender Elements .................................................................... E-42
Example E.9 Rectangular HSS Compression Member without Slender Elements ......................................... E-47
Example E.10 Rectangular HSS Compression Member with Slender Elements ............................................... E-50
Example E.11 Pipe Compression Member ........................................................................................................ E-54
Example E.12 Built-up I-Shaped Member with Different Flange Sizes ............................................................ E-57
Example E.13 Double WT Compression Member............................................................................................. E-63
CHAPTER F DESIGN OF MEMBERS FOR FLEXURE.............................................................................. F-1
Example F.1-1A W-Shape Flexural Member Design in Strong-Axis Bending, Continuously Braced ................... F-6
Example F.1-1B W-Shape Flexural Member Design in Strong-Axis Bending, Continuously Braced................... F-8
Example F.1-2A W-Shape Flexural Member Design in Strong-Axis Bending, Braced at Third Points ................. F-9
Example F.1-2B W-Shape Flexural Member Design in Strong-Axis Bending, Braced at Third Points............... F-10

vi
Example F.1-3A W-Shape Flexural Member Design in Strong-Axis Bending, Braced at Midspan..................... F-12
Example F.1-3B W-Shape Flexural Member Design in Strong-Axis Bending, Braced at Midspan..................... F-14
Example F.2-1A Compact Channel Flexural Member, Continuously Braced ...................................................... F-16
Example F.2-1B Compact Channel Flexural Member, Continuously Braced ...................................................... F-18
Example F.2-2A Compact Channel Flexural Member with Bracing at Ends and Fifth Points ............................. F-19
Example F.2-2B Compact Channel Flexural Member with Bracing at End and Fifth Points ............................... F-20
Example F.3A W-Shape Flexural Member with Noncompact Flanges in Strong-Axis Bending ...................... F-22
Example F.3B W-Shape Flexural Member with Noncompact Flanges in Strong-Axis Bending ...................... F-24
Example F.4 W-Shape Flexural Member, Selection by Moment of Inertia for Strong-Axis Bending ............ F-26
Example F.5 I-Shaped Flexural Member in Minor-Axis Bending .................................................................. F-28
Example F.6 HSS Flexural Member with Compact Flanges ........................................................................... F-30
Example F.7A HSS Flexural Member with Noncompact Flanges ..................................................................... F-32
Example F.7B HSS Flexural Member with Noncompact Flanges ..................................................................... F-34
Example F.8A HSS Flexural Member with Slender Flanges ............................................................................. F-36
Example F.8B HSS Flexural Member with Slender Flanges ............................................................................. F-38
Example F.9A Pipe Flexural Member ................................................................................................................ F-41
Example F.9B Pipe Flexural Member ................................................................................................................ F-42
Example F.10 WT-Shape Flexural Member ..................................................................................................... F-44
Example F.11A Single Angle Flexural Member .................................................................................................. F-47
Example F.11B Single Angle Flexural Member .................................................................................................. F-50
Example F.11C Single Angle Flexural Member .................................................................................................. F-53
Example F.12 Rectangular Bar in Strong-Axis Bending .................................................................................. F-59
Example F.13 Round Bar in Bending ............................................................................................................... F-61
Example F.14 Point-Symmetrical Z-shape in Strong-Axis Bending................................................................. F-63
Chapter F
Design Example
References .................................................................................................................................................... F-69
CHAPTER G DESIGN OF MEMBERS FOR SHEAR...................................................................................G-1
Example G.1A W-Shape in Strong-Axis Shear ....................................................................................................G-3
Example G.1B W-Shape in Strong-Axis Shear ....................................................................................................G-4
Example G.2A C-Shape in Strong-Axis Shear .....................................................................................................G-5
Example G.2B C-Shape in Strong-Axis Shear .....................................................................................................G-6
Example G.3 Angle in Shear .............................................................................................................................G-7
Example G.4 Rectangular HSS in Shear ............................................................................................................G-9
Example G.5 Round HSS in Shear ..................................................................................................................G-11
Example G.6 Doubly Symmetric Shape in Weak-Axis Shear .........................................................................G-13
Example G.7 Singly Symmetric Shape in Weak-Axis Shear ...........................................................................G-15
Example G.8A Built-up Girder with Transverse Stiffeners ................................................................................G-17
Example G.8B Built-up Girder with Transverse Stiffeners ................................................................................G-21
Chapter G
Design Example
References ....................................................................................................................................................G-24
CHAPTER H DESIGN OF MEMBERS FOR COMBINED FORCES AND TORSION ............................H-1
Example H.1A W-shape Subject to Combined Compression and Bending
About Both Axes (Braced Frame) ...............................................................................................H-2
Example H.1B W-shape Subject to Combined Compression and Bending Moment
About Both Axes (Braced Frame) ................................................................................................H-4
Example H.2 W-Shape Subject to Combined Compression and Bending Moment
About Both Axes (By AISC Specification Section H2) ..............................................................H-5
Example H.3 W-Shape Subject to Combined Axial Tension and Flexure......................................................... H-8
Example H.4 W-Shape Subject to Combined Axial Compression and Flexure ..............................................H-12
Example H.5A Rectangular HSS Torsional Strength .........................................................................................H-16

vii
Example H.5B Round HSS Torsional Strength ..................................................................................................H-17
Example H.5C HSS Combined Torsional and Flexural Strength .......................................................................H-19
Example H.6 W-Shape Torsional Strength ......................................................................................................H-23
Chapter H
Design Example
References ....................................................................................................................................................H-30
CHAPTER I DESIGN OF COMPOSITE MEMBERS...................................................................................I-1
Example I.1 Composite Beam Design ............................................................................................................... I-8
Example I.2 Composite Girder Design ........................................................................................................... I-18
Example I.3 Concrete Filled Tube (CFT) Force Allocation and Load Transfer .............................................. I-35
Example I.4 Concrete Filled Tube (CFT) in Axial Compression .................................................................... I-45
Example I.5 Concrete Filled Tube (CFT) in Axial Tension ............................................................................ I-50
Example I.6 Concrete Filled Tube (CFT) in Combined Axial Compression, Flexure and Shear ................... I-52
Example I.7 Concrete Filled Box Column with Noncompact/Slender Elements ............................................ I-66
Example I.8 Encased Composite Member Force Allocation and Load Transfer ............................................ I-80
Example I.9 Encased Composite Member in Axial Compression ................................................................... I-93
Example I.10 Encased Composite Member in Axial Tension ......................................................................... I-100
Example I.11 Encased Composite Member in Combined Axial Compression, Flexure and Shear ................ I-103
Example I.12 Steel Anchors in Composite Components ................................................................................. I-119
Chapter I
Design Example
References ................................................................................................................................................... I-123
CHAPTER J DESIGN OF CONNECTIONS...................................................................................................J-1
Example J.1 Fillet Weld in Longitudinal Shear .................................................................................................J-2
Example J.2 Fillet Weld Loaded at an Angle ....................................................................................................J-4
Example J.3 Combined Tension and Shear in Bearing Type Connections........................................................ J-6
Example J.4A Slip-Critical Connection with Short-Slotted Holes .......................................................................J-8
Example J.4B Slip-Critical Connection with Long-Slotted Holes .................................................J-10
Example J.5 Combined Tension and Shear in a Slip-Critical Connection .................................................J-12
Example J.6 Bearing Strength of a Pin in a Drilled Hole ................................................................................ J-15
Example J.7 Base Plate Bearing on Concrete...................................................................................................J-16
CHAPTER K DESIGN OF HSS AND BOX MEMBER CONNECTIONS...................................................K-1
Example K.1 Welded/Bolted Wide Tee Connection to an HSS Column ...........................................................K-2
Example K.2 Welded/Bolted Narrow Tee Connection to an HSS Column .....................................................K-10
Example K.3 Double Angle Connection to an HSS Column ...........................................................................K-13
Example K.4 Unstiffened Seated Connection to an HSS Column ...................................................................K-16
Example K.5 Stiffened Seated Connection to an HSS Column .......................................................................K-19
Example K.6 Single-Plate Connection to Rectangular HSS Column ..............................................................K-24
Example K.7 Through-Plate Connection .........................................................................................................K-27
Example K.8 Transverse Plate Loaded Perpendicular to the HSS Axis on a Rectangular HSS .......................K-31
Example K.9 Longitudinal Plate Loaded Perpendicular to the HSS Axis on a Round HSS ............................K-34
Example K.10 HSS Brace Connection to a W-Shape Column...........................................................................K-36
Example K.11 Rectangular HSS Column with a Cap Plate, Supporting a Continuous Beam ...........................K-39
Example K.12 Rectangular HSS Column Base Plate ........................................................................................K-42
Example K.13 Rectangular HSS Strut End Plate ...............................................................................................K-45
Chapter K
Design Example
References ....................................................................................................................................................K-49

viii
PART II. EXAMPLES BASED ON THE AISC STEEL CONSTRUCTION MANUAL
CHAPTER IIA SIMPLE SHEAR CONNECTIONS.......................................................................................IIA-1
Example II.A-1 All-Bolted Double-Angle Connection ...................................................................................... IIA-2
Example II.A-2 Bolted/Welded Double-Angle Connection ............................................................................... IIA-4
Example II.A-3 All-Welded Double-Angle Connection .................................................................................... IIA-6
Example II.A-4 All-Bolted Double-Angle Connection in a Coped Beam .......................................................... IIA-9
Example II.A-5 Welded/ Bolted Double-Angle Connection (Beam-to-Girder Web). ...................................... IIA-13
Example II.A-6 Beam End Coped at the Top Flange Only .............................................................................. IIA-16
Example II.A-7 Beam End Coped at the Top and Bottom Flanges.................................................................. IIA-23
Example II.A-8 All-Bolted Double-Angle Connections (Beams-to-Girder Web) ........................................... IIA-26
Example II.A-9 Offset All-Bolted Double-Angle Connections (Beams-to-Girder Web) ................................ IIA-34
Example II.A-10 Skewed Double Bent-Plate Connection (Beam-to-Girder Web)............................................. IIA-37
Example II.A-11 Shear End-Plate Connection (Beam to Girder Web). ............................................................. IIA-43
Example II.A-12 All-Bolted Unstiffened Seated Connection (Beam-to-Column Web) ..................................... IIA-45
Example II.A-13 Bolted/Welded Unstiffened Seated Connection (Beam-to-Column Flange) .......................... IIA-48
Example II.A-14 Stiffened Seated Connection (Beam-to-Column Flange) ........................................................ IIA-51
Example II.A-15 Stiffened Seated Connection (Beam-to-Column Web) ........................................................... IIA-54
Example II.A-16 Offset Unstiffened Seated Connection (Beam-to-Column Flange)......................................... IIA-57
Example II.A-17 Single-Plate Connection (Conventional – Beam-to-Column Flange) ..................................... IIA-60
Example II.A-18 Single-Plate Connection (Beam-to-Girder Web) .................................................................... IIA-62
Example II.A-19 Extended Single-Plate Connection (Beam-to-Column Web) .................................................. IIA-64
Example II.A-20 All-Bolted Single-Plate Shear Splice ...................................................................................... IIA-70
Example II.A-21 Bolted/Welded Single-Plate Shear Splice ............................................................................... IIA-75
Example II.A-22 Bolted Bracket Plate Design ................................................................................................... IIA-80
Example II.A-23 Welded Bracket Plate Design. ................................................................................................ IIA-86
Example II.A-24 Eccentrically Loaded Bolt Group (IC Method) ....................................................................... IIA-91
Example II.A-25 Eccentrically Loaded Bolt Group (Elastic Method)................................................................ IIA-93
Example II.A-26 Eccentrically Loaded Weld Group (IC Method)..................................................................... IIA-95
Example II.A-27 Eccentrically Loaded Weld Group (Elastic Method) .............................................................. IIA-98
Example II.A-28 All-Bolted Single-Angle Connection (Beam-to-Girder Web) .............................................. IIA-100
Example II.A-29 Bolted/Welded Single-Angle Connection (Beam-to-Column Flange).................................. IIA-105
Example II.A-30 All-Bolted Tee Connection (Beam-to-Column Flange) ........................................................ IIA-108
Example II.A-31 Bolted/Welded Tee Connection (Beam-to-Column Flange) ................................................. IIA-115
CHAPTER IIB FULLY RESTRAINED (FR) MOMENT CONNECTIONS............................................... IIB-1
Example II.B-1 Bolted Flange-Plate FR Moment Connection (Beam-to-Column Flange) ............................... IIB-2
Example II.B-2 Welded Flange-Plated FR Moment Connection (Beam-to-Column Flange) ......................... IIB-14
Example II.B-3 Directly Welded Flange FR Moment Connection (Beam-to-Column Flange). ..................... IIB-20
Example II.B-4 Four-Bolt Unstiffened Extended End-Plate
FR Moment Connection (Beam-to-Column Flange) ............................................................. IIB-22
Chapter IIB
Design Example
References ................................................................................................................................................. IIB-33
CHAPTER IIC BRACING AND TRUSS CONNECTIONS.......................................................................... IIC-1
Example II.C-1 Truss Support Connection........................................................................................................ IIC-2
Example II.C-2 Bracing Connection ............................................................................................................... IIC-13
Example II.C-3 Bracing Connection ............................................................................................................... IIC-41
Example II.C-4 Truss Support Connection...................................................................................................... IIC-50
Example II.C-5 HSS Chevron Brace Connection ............................................................................................ IIC-57
Example II.C-6 Heavy Wide Flange Compression Connection (Flanges on the Outside) .............................. IIC-69
CHAPTER IID MISCELLANEOUS CONNECTIONS.................................................................................. ID-1

ix
Example II.D-1 Prying Action in Tees and in Single Angles ............................................................................ IID-2
Example II.D-2 Beam Bearing Plate ................................................................................................................. IID-9
Example II.D-3 Slip-Critical Connection with Oversized Holes...................................................................... IID-15
PART III. SYSTEM DESIGN EXAMPLES ................................................................. III-1
Example III-1 Design of Selected Members and Lateral Analysis of a Four-Story Building .............................III-2
Introduction..................................................................................................................................III-2
Conventions .................................................................................................................................III-2
Design Sequence..........................................................................................................................III-3
General Description of the Building ............................................................................................III-4
Roof Member Design and Selection ...........................................................................................III-5
Floor Member Design and Selection ........................................................................................III-17
Column Design and Selection for Gravity Loads ..................................................................... III-38
Wind Load Determination ........................................................................................................III-46
Seismic Load Determination .....................................................................................................III-49
Moment Frame Model ...............................................................................................................III-61
Calculation of Required Strength—Three Methods ..................................................................III-67
Beam Analysis in the Moment Frame........................................................................................III-77
Braced Frame Analysis..............................................................................................................III-80
Analysis of Drag Struts..............................................................................................................III-84
Part III Example
References ...................................................................................................................................................III-87
PART IV. ADDITIONAL RESOURCES.....................................................................IV-1
Table 4-1 Discussion................................................................................................................................... IV-2
Table 4-1 W-Shapes in Axial Compressions, Fy = 65 ksi ........................................................................... IV-5
Table 6-1 Discussion................................................................................................................................. IV-17
Table 6-1 Combined Flexure and Axial Force, W-Shapes, Fy = 65 ksi .................................................... IV-19

A-1
Chapter A
General Provisions
A1. SCOPE
These design examples are intended to illustrate the application of the 2010 AISC Specification for Structural
Steel Buildings (ANSI/AISC 360-10) (AISC, 2010) and the AISC Steel Construction Manual, 14th Edition
(AISC, 2011) in low-seismic applications. For information on design applications requiring seismic detailing, see
the AISC Seismic Design Manual.
A2. REFERENCED SPECIFICATIONS, CODES AND STANDARDS
Section A2 includes a detailed list of the specifications, codes and standards referenced throughout the AISC
Specification.
A3. MATERIAL
Section A3 includes a list of the steel materials that are approved for use with the AISC Specification. The
complete ASTM standards for the most commonly used steel materials can be found in Selected ASTM Standards
for Structural Steel Fabrication (ASTM, 2011).
A4. STRUCTURAL DESIGN DRAWINGS AND SPECIFICATIONS
Section A4 requires that structural design drawings and specifications meet the requirements in the AISC Code of
Standard Practice for Steel Buildings and Bridges (AISC, 2010b).
Return to Table of Contents

A-2
CHAPTER A REFERENCES
AISC (2010a), Specification for Structural Steel Buildings, ANSI/AISC 360-10, American Institute for Steel
Construction, Chicago, IL.
AISC (2010b), Code of Standard Practice for Steel Buildings and Bridges, American Institute for Steel
Construction, Chicago, IL.
AISC (2011), Steel Construction Manual, 14th Ed., American Institute for Steel Construction, Chicago, IL.
ASTM (2011), Selected ASTM Standards for Structural Steel Fabrication, ASTM International, West
Conshohocken, PA.

B-1
Chapter B
Design Requirements
B1. GENERAL PROVISIONS
B2. LOADS AND LOAD COMBINATIONS
In the absence of an applicable building code, the default load combinations to be used with this Specification are
those from Minimum Design Loads for Buildings and Other Structures (ASCE/SEI 7-10) (ASCE, 2010).
B3. DESIGN BASIS
Chapter B of the AISC Specification and Part 2 of the AISC Manual describe the basis of design, for both LRFD
and ASD.
This Section describes three basic types of connections: simple connections, fully restrained (FR) moment
connections, and partially restrained (PR) moment connections. Several examples of the design of each of these
types of connection are given in Part II of these design examples.
Information on the application of serviceability and ponding provisions may be found in AISC Specification
Chapter L and AISC Specification Appendix 2, respectively, and their associated commentaries. Design examples
and other useful information on this topic are given in AISC Design Guide 3, Serviceability Design
Considerations for Steel Buildings, Second Edition (West et al., 2003).
Information on the application of fire design provisions may be found in AISC Specification Appendix 4 and its
associated commentary. Design examples and other useful information on this topic are presented in AISC Design
Guide 19, Fire Resistance of Structural Steel Framing (Ruddy et al., 2003).
Corrosion protection and fastener compatibility are discussed in Part 2 of the AISC Manual.
B4. MEMBER PROPERTIES
AISC Specification Tables B4.1a and B4.1b give the complete list of limiting width-to-thickness ratios for all
compression and flexural members defined by the AISC Specification.
Except for one section, the W-shapes presented in the compression member selection tables as column sections
meet the criteria as nonslender element sections. The W-shapes presented in the flexural member selection tables
as beam sections meet the criteria for compact sections, except for 10 specific shapes. When noncompact or
slender-element sections are tabulated in the design aids, local buckling criteria are accounted for in the tabulated
design values.
The shapes listing and other member design tables in the AISC Manual also include footnoting to highlight
sections that exceed local buckling limits in their most commonly available material grades. These footnotes
include the following notations:
c
Shape is slender for compression.
f
Shape exceeds compact limit for flexure.
g
The actual size, combination and orientation of fastener components should be compared with the geometry of
the cross section to ensure compatibility.
h
Flange thickness greater than 2 in. Special requirements may apply per AISC Specification Section A3.1c.
v
Shape does not meet the h/tw limit for shear in AISC Specification Section G2.1a.

B-2
CHAPTER B REFERENCES
ASCE (2010), Minimum Design Loads for Buildings and Other Structures, ASCE/SEI 7-10, American Society of
Civil Engineers, Reston, VA.
West, M., Fisher, J. and Griffis, L.G. (2003), Serviceability Design Considerations for Steel Buildings, Design
Guide 3, 2nd Ed., AISC, Chicago, IL.
Ruddy, J.L., Marlo, J.P., Ioannides, S.A. and Alfawakhiri, F. (2003), Fire Resistance of Structural Steel Framing,
Design Guide 19, AISC, Chicago, IL.

C-1
Chapter C
Design for Stability
C1. GENERAL STABILITY REQUIREMENTS
The AISC Specification requires that the designer account for both the stability of the structural system as a
whole, and the stability of individual elements. Thus, the lateral analysis used to assess stability must include
consideration of the combined effect of gravity and lateral loads, as well as member inelasticity, out-of-
plumbness, out-of-straightness and the resulting second-order effects, P-Δ and P-δ. The effects of “leaning
columns” must also be considered, as illustrated in the examples in this chapter and in the four-story building
design example in Part III of AISC Design Examples.
P-Δ and P-δ effects are illustrated in AISC Specification Commentary Figure C-C2.1. Methods for addressing
stability, including P-Δ and P-δ effects, are provided in AISC Specification Section C2 and Appendix 7.
C2. CALCULATION OF REQUIRED STRENGTHS
The calculation of required strengths is illustrated in the examples in this chapter and in the four-story building
design example in Part III of AISC Design Examples.
C3. CALCULATION OF AVAILABLE STRENGTHS
The calculation of available strengths is illustrated in the four-story building design example in Part III of AISC
Design Examples.

C-2
EXAMPLE C.1A DESIGN OF A MOMENT FRAME BY THE DIRECT ANALYSIS METHOD
Given:
Determine the required strengths and effective length factors for the columns in the rigid frame shown below for
the maximum gravity load combination, using LRFD and ASD. Use the direct analysis method. All members are
ASTM A992 material.
Columns are unbraced between the footings and roof in the x- and y-axes and are assumed to have pinned bases.
Solution:
From Manual Table 1-1, the W12×65 has A = 19.1 in.2
The beams from grid lines A to B, and C to E and the columns at A, D and E are pinned at both ends and do not
contribute to the lateral stability of the frame. There are no P-Δ effects to consider in these members and they may
be designed using K=1.0.
The moment frame between grid lines B and C is the source of lateral stability and therefore must be designed
using the provisions of Chapter C of the AISC Specification. Although the columns at grid lines A, D and E do
not contribute to lateral stability, the forces required to stabilize them must be considered in the analysis. For the
analysis, the entire frame could be modeled or the model can be simplified as shown in the figure below, in which
the stability loads from the three “leaning” columns are combined into a single column.
From Chapter 2 of ASCE/SEI 7, the maximum gravity load combinations are:
LRFD ASD
wu = 1.2D + 1.6L
= 1.2(0.400 kip/ft) + 1.6(1.20 kip/ft)
= 2.40 kip/ft
wa = D + L
= 0.400 kip/ft + 1.20 kip/ft
= 1.60 kip/ft
Per AISC Specification Section C2.1, for LRFD perform a second-order analysis and member strength checks
using the LRFD load combinations. For ASD, perform a second-order analysis using 1.6 times the ASD load
combinations and divide the analysis results by 1.6 for the ASD member strength checks.
Frame Analysis Gravity Loads
The uniform gravity loads to be considered in a second-order analysis on the beam from B to C are:

C-3
LRFD ASD
wu' = 2.40 kip/ft wa' = 1.6(1.60 kip/ft)
= 2.56 kip/ft
Concentrated gravity loads to be considered in a second-order analysis on the columns at B and C contributed by
adjacent beams are:
LRFD ASD
Pu'= (15.0 ft)(2.40 kip/ft)
= 36.0 kips
Pa' = 1.6(15.0 ft)(1.60 kip/ft)
= 38.4 kips
Concentrated Gravity Loads on the Pseudo “Leaning” Column
The load in this column accounts for all gravity loading that is stabilized by the moment frame, but is not directly
applied to it.
LRFD ASD
PuL' = (60.0 ft)(2.40 kip/ft)
= 144 kips
PaL' = 1.6(60.0 ft)(1.60 kip/ft)
= 154 kips
Frame Analysis Notional Loads
Per AISC Specification Section C2.2, frame out-of-plumbness must be accounted for either by explicit modeling
of the assumed out-of-plumbness or by the application of notional loads. Use notional loads.
From AISC Specification Equation C2-1, the notional loads are:
LRFD ASD
α = 1.0
Yi = (120 ft)(2.40 kip/ft)
= 288 kips
Ni = 0.002αYi (Spec. Eq. C2-1)
= 0.002(1.0)(288 kips)
= 0.576 kips
α = 1.6
Yi = (120 ft)(1.60 kip/ft)
= 192 kips
= 0.002(1.6)(192 kips)
= 0.614 kips
Summary of Applied Frame Loads
LRFD ASD

C-4
Per AISC Specification Section C2.3, conduct the analysis using 80% of the nominal stiffnesses to account for the
effects of inelasticity. Assume, subject to verification, that αPr /Py is no greater than 0.5; therefore, no additional
stiffness reduction is required.
50% of the gravity load is carried by the columns of the moment resisting frame. Because the gravity load
supported by the moment resisting frame columns exceeds one third of the total gravity load tributary to the
frame, per AISC Specification Section C2.1, the effects of P-δ upon P-Δ must be included in the frame analysis. If
the software used does not account for P-δ effects in the frame analysis, this may be accomplished by adding
joints to the columns between the footing and beam.
Using analysis software that accounts for both P-Δ and P-δ effects, the following results are obtained:
First-order results
LRFD ASD
Δ1st = 0.149 in. Δ1st = 0.159 in. (prior to dividing by 1.6)
Second-order results
LRFD ASD
Δ2nd = 0.217 in.
2
1
0.217 in.
0.149 in.
nd
st
Δ
=
Δ
= 1.46
Δ2nd = 0.239 in. (prior to dividing by 1.6)
2
1
0.239 in.
0.159 in.
nd
st
Δ
=
Δ
= 1.50
Check the assumption that 0.5r yP Pα ≤ and therefore, τb = 1.0:

C-5
Py = FyAg
= 50 ksi(19.1 in.2
)
= 955 kips
LRFD ASD
( )1.0 72.6kips
955kips
r
y
P
P
α
=
0.0760 0.5= ≤ o.k.
( )1.6 48.4kips
955kips
r
y
P
P
α
=
0.0811 0.5= ≤ o.k.
The stiffness assumption used in the analysis, τb = 1.0, is verified.
Although the second-order sway multiplier is approximately 1.5, the change in bending moment is small because
the only sway moments are those produced by the small notional loads. For load combinations with significant
gravity and lateral loadings, the increase in bending moments is larger.
Verify the column strengths using the second-order forces shown above, using the following effective lengths
(calculations not shown):
Columns:
Use KLx = 20.0 ft
Use KLy = 20.0 ft

C-6
EXAMPLE C.1B DESIGN OF A MOMENT FRAME BY THE EFFECTIVE LENGTH METHOD
Repeat Example C.1A using the effective length method.
Given:
the maximum gravity load combination, using LRFD and ASD. Use the effective length method.
Solution:
From Manual Table 1-1, the W12×65 has Ix = 533 in.4
using the provisions of Appendix 7 of the AISC Specification. Although the columns at grid lines A, D and E do
not contribute to lateral stability, the forces required to stabilize them must be considered in the analysis. For the
analysis, the entire frame could be modeled or the model can be simplified as shown in the figure below, in which
the stability loads from the three “leaning” columns are combined into a single column.
Check the limitations for the use of the effective length method given in Appendix 7, Section 7.2.1:
(1) The structure supports gravity loads through nominally vertical columns.
(2) The ratio of maximum second-order drift to the maximum first-order drift will be assumed to be no
greater than 1.5, subject to verification following.
LRFD ASD
wu = 1.2D + 1.6L
= 1.2(0.400 kip/ft) + 1.6(1.20 kip/ft)
= 2.40 kip/ft
wa = D + L
= 0.400 kip/ft + 1.20 kip/ft
= 1.60 kip/ft
Per AISC Specification Appendix 7, Section 7.2.1, the analysis must conform to the requirements of AISC
Specification Section C2.1, with the exception of the stiffness reduction required by the provisions of Section
C2.3.

C-7
Per AISC Specification Section C2.1, for LRFD perform a second-order analysis and member strength checks
using the LRFD load combinations. For ASD, perform a second-order analysis at 1.6 times the ASD load
combinations and divide the analysis results by 1.6 for the ASD member strength checks.
The uniform gravity loads to be considered in a second-order analysis on the beam from B to C are:
LRFD ASD
wu' = 2.40 kip/ft wa' = 1.6(1.60 kip/ft)
= 2.56 kip/ft
adjacent beams are:
LRFD ASD
Pu'= (15.0 ft)(2.40 kip/ft)
= 36.0 kips
Pa' = 1.6(15.0 ft)(1.60 kip/ft)
= 38.4 kips
Concentrated Gravity Loads on the Pseudo “Leaning” Column
The load in this column accounts for all gravity loading that is stabilized by the moment frame, but is not directly
applied to it.
LRFD ASD
PuL' = (60.0 ft)(2.40 kip/ft)
= 144 kips
PaL'= 1.6(60.0 ft)(1.60 kip/ft)
= 154 kips
Per AISC Specification Appendix 7, Section 7.2.2, frame out-of-plumbness must be accounted for by the
application of notional loads in accordance with AISC Specification Section C2.2b.
From AISC Specification Equation C2-1, the notional loads are:
LRFD ASD
α = 1.0
Yi = (120 ft)(2.40 kip/ft)
= 288 kips
= 0.002(1.0)(288 kips)
= 0.576 kips
α = 1.6
Yi = (120 ft)(1.60 kip/ft)
= 192 kips
= 0.002(1.6)(192 kips)
= 0.614 kips

C-8
LRFD ASD
Per AISC Specification Appendix 7, Section 7.2.2, conduct the analysis using the full nominal stiffnesses.
50% of the gravity load is carried by the columns of the moment resisting frame. Because the gravity load
supported by the moment resisting frame columns exceeds one third of the total gravity load tributary to the
frame, per AISC Specification Section C2.1, the effects of P-δ must be included in the frame analysis. If the
software used does not account for P-δ effects in the frame analysis, this may be accomplished by adding joints to
the columns between the footing and beam.
Using analysis software that accounts for both P-Δ and P-δ effects, the following results are obtained:
First-order results
LRFD ASD
Δ1st = 0.119 in. Δ1st = 0.127 in. (prior to dividing by 1.6)

C-9
Second-order results
LRFD ASD
Δ2nd = 0.159 in.
2
1
0.159 in.
0.119 in.
nd
st
Δ
=
Δ
= 1.34
Δ2nd = 0.174 in. (prior to dividing by 1.6)
2
1
0.174 in.
0.127 in.
nd
st
Δ
=
Δ
= 1.37
The assumption that the ratio of the maximum second-order drift to the maximum first-order drift is no greater
than 1.5 is verified; therefore, the effective length method is permitted.
the only sway moments for this load combination are those produced by the small notional loads. For load
combinations with significant gravity and lateral loadings, the increase in bending moments is larger.
Calculate the in-plane effective length factor, Kx, using the “story stiffness method” and Equation C-A-7-5
presented in Commentary Appendix 7, Section 7.2. Take Kx = K2
( )
2 2
2 2 2
0.85 0.15 1.7
r H H
x
L r
P EI EI
K K
R P HL HLL L
⎛ ⎞Σ π Δ π Δ⎛ ⎞ ⎛ ⎞
= = ≥⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟+ Σ⎝ ⎠ ⎝ ⎠⎝ ⎠
(Spec. Eq. C-A-7-5)
Calculate the total load in all columns, rPΣ
LRFD ASD
( )2.40 kip/ft 120 ft
288 kips
rPΣ =
=
( )1.60 kip/ft 120 ft
192 kips
rPΣ =
=
Calculate the ratio of the leaning column loads to the total load, RL
LRFD ASD
( )288 kips 71.5 kips 72.5 kips
288 kips
0.500
r r moment frame
L
r
P P
R
P
Σ − Σ
=
Σ
− +
=
=
( )192 kips 47.7 kips 48.3 kips
192 kips
0.500
r r moment frame
L
r
P P
R
P
Σ − Σ
=
Σ
− +
=
=
Calculate the Euler buckling strength of an individual column.

C-10
( )( )
( )
2 42
2 2
29,000 ksi 533 in.
240 in.
2,650 kips
EI
L
ππ
=
=
Calculate the drift ratio using the first-order notional loading results.
LRFD ASD
0.119in.
240in.
0.000496in./in.
H
L
Δ
=
=
0.127in.
240in.
0.000529in./in.
H
L
Δ
=
=
For the column at line C:
LRFD ASD
( ) ( )
( )
( )
288 kips
0.85 0.15 0.500 72.4 kips
0.000496 in./in.
2,650 kips
0.576 kips
0.000496 in./in.
2,650 kips
1.7 7.35 kips
3.13 0.324
xK
+⎡ ⎤⎣ ⎦
=
⎛ ⎞
× ⎜ ⎟
⎝ ⎠
⎛ ⎞
≥ ⎜ ⎟⎜ ⎟
⎝ ⎠
= ≥
Use Kx = 3.13
( )
( ) ( )( )
( )
( )( )
1.6 192 kips
0.85 0.15 0.500 1.6 48.3 kips
0.000529 in./in.
2,650 kips
0.614 kips
0.000529 in./in.
2,650 kips
1.7 1.6 4.90 kips
3.13 0.324
xK
+⎡ ⎤⎣ ⎦
=
⎛ ⎞
× ⎜ ⎟
⎝ ⎠
⎛ ⎞
≥ ⎜ ⎟⎜ ⎟
⎝ ⎠
= ≥
Use Kx = 3.13
Note that it is necessary to multiply the column loads by 1.6 for ASD in the expression above.
Verify the column strengths using the second-order forces shown above, using the following effective lengths
(calculations not shown):
Columns:
Use KxLx = ( )3.13 20.0ft 62.6ft=
Use KyLy = 20.0 ft

C-11
EXAMPLE C.1C DESIGN OF A MOMENT FRAME BY THE FIRST-ORDER METHOD
Repeat Example C.1A using the first-order analysis method.
Given:
the maximum gravity load combination, using LRFD and ASD. Use the first-order analysis method.
Solution:
From Manual Table 1-1, the W12×65 has A = 19.1 in.2
using the provisions of Appendix 7 of the AISC Specification. Although the columns at grid lines A, D and E do
not contribute to lateral stability, the forces required to stabilize them must be considered in the analysis. These
members need not be included in the analysis model, except that the forces in the “leaning” columns must be
included in the calculation of notional loads.
Check the limitations for the use of the first-order analysis method given in Appendix 7, Section 7.3.1:
(1) The structure supports gravity loads through nominally vertical columns.
(2) The ratio of maximum second-order drift to the maximum first-order drift will be assumed to be no
greater than 1.5, subject to verification.
(3) The required axial strength of the members in the moment frame will be assumed to be no more than
50% of the axial yield strength, subject to verification.
LRFD ASD
wu = 1.2D + 1.6L
= 1.2(0.400 kip/ft) + 1.6(1.20 kip/ft)
= 2.40 kip/ft
wa = D + L
= 0.400 kip/ft + 1.20 kip/ft
= 1.60 kip/ft

C-12
Per AISC Specification Appendix 7, Section 7.3.2, the required strengths are determined from a first-order
analysis using notional loads determined below along with a B1 multiplier as determined from Appendix 8.
For ASD, do not multiply loads or divide results by 1.6.
The uniform gravity loads to be considered in the first-order analysis on the beam from B to C are:
LRFD ASD
wu' = 2.40 kip/ft wa' = 1.60 kip/ft
adjacent beams are:
LRFD ASD
Pu'= (15.0 ft)(2.40 kip/ft)
= 36.0 kips
Pa' = (15.0 ft)(1.60 kip/ft)
= 24.0 kips
Per AISC Specification Appendix 7, Section 7.3.2, frame out-of-plumbness must be accounted for by the
application of notional loads.
From AISC Specification Appendix Equation A-7-2, the required notional loads are:
LRFD ASD
α = 1.0
Yi = (120 ft)(2.40 kip/ft)
= 288 kips
Δ = 0.0 in. (no drift in this load combination)
L = 240 in.
Ni = 2.1α(Δ/L)Yi ≥ 0.0042Yi
= 2.1(1.0)(0.0 in./240 in.)(288 kips)
≥ 0.0042(288 kips)
= 0.0 kips ≥ 1.21 kips
Use Ni = 1.21 kips
α = 1.6
Yi = (120 ft)(1.60 kip/ft)
= 192 kips
Δ = 0.0 in. (no drift in this load combination)
L = 240 in.
Ni = 2.1α(Δ/L)Yi ≥ 0.0042Yi
= 2.1(1.6)(0.0 in./240 in.)(192 kips)
≥ 0.0042(192 kips)
= 0.0 kips ≥ 0.806 kips
Use Ni = 0.806 kips

C-13
LRFD ASD
Per AISC Specification Appendix 7, Section 7.2.2, conduct the analysis using the full nominal stiffnesses.
Using analysis software, the following first-order results are obtained:
LRFD ASD
Δ1st = 0.250 in. Δ1st = 0.167 in.
Check the assumption that the ratio of the second-order drift to the first-order drift does not exceed 1.5. B2 can be
used to check this limit. Calculate B2 per the provisions of Section 8.2.2 of Appendix 8 using the results of the
first-order analysis.
LRFD ASD
Pmf = 71.2 kips + 72.8 kips
= 144 kips
Pstory = 144 kips + 4(36 kips)
= 288 kips
( )1 0.15M mf storyR P P= − (Spec. Eq. A-8-8)
( )1 0.15 144kips 288kips
0.925
= −
=
ΔH = 0.250 in.
H = 1.21 kips
L = 240 in.
Pmf = 47.5 kips + 48.5 kips
= 96 kips
Pstory = 96 kips + 4(24 kips)
= 192 kips
( )1 0.15M mf storyR P P= − (Spec. Eq. A-8-8)
( )1 0.15 96kips 192kips
0.925
= −
=
ΔH = 0.167 in.
H = 0.806 kips
L = 240 in.

C-14
LRFD ASD
estory M
H
HL
P R=
Δ
(Spec. Eq. A-8-7)
( )1.21 kips 240 in.
0.925
0.250 in.
1,070 kips
=
=
α = 1.00
2
1
1
1
story
e story
B
P
P
= ≥
α
−
(Spec. Eq. A-8-6)
( )
1
1
1.00 288 kips
1
1,070 kips
1.37
= ≥
−
=
estory M
H
HL
P R=
Δ
(Spec. Eq. A-8-7)
( )0.806 kips 240 in.
0.925
0.167 in.
1,070 kips
=
=
α = 1.60
2
1
1
1
story
e story
B
P
P
= ≥
α
−
(Spec. Eq. A-8-6)
( )
1
1
1.60 192 kips
1
1,070 kips
1.40
= ≥
−
=
The assumption that the ratio of the maximum second-order drift to the maximum first-order drift is no greater
than 1.5 is correct; therefore, the first-order analysis method is permitted.
Check the assumption that 0.5r yP Pα ≤ and therefore, the first-order analysis method is permitted.
0.5Py = 0.5FyAg
= 0.5(50 ksi)(19.1 in.2
)
= 478 kips
LRFD ASD
( )1.0 72.8 kipsrPα =
72.8 kips 478 kips= < o.k.
( )1.6 48.5 kipsrPα =
77.6 kips 478 kips= < o.k.
The assumption that the first-order analysis method can be used is verified.
the only sway moments are those produced by the small notional loads. For load combinations with significant
gravity and lateral loadings, the increase in bending moments is larger.
Verify the column strengths using the second-order forces, using the following effective lengths (calculations not
shown):
Columns:
Use KLx = 20.0 ft
Use KLy = 20.0 ft

D-1
Chapter D
Design of Members for Tension
D1. SLENDERNESS LIMITATIONS
Section D1 does not establish a slenderness limit for tension members, but recommends limiting L/r to a
maximum of 300. This is not an absolute requirement. Rods and hangers are specifically excluded from this
recommendation.
D2. TENSILE STRENGTH
Both tensile yielding strength and tensile rupture strength must be considered for the design of tension members.
It is not unusual for tensile rupture strength to govern the design of a tension member, particularly for small
members with holes or heavier sections with multiple rows of holes.
For preliminary design, tables are provided in Part 5 of the AISC Manual for W-shapes, L-shapes, WT-shapes,
rectangular HSS, square HSS, round HSS, Pipe and 2L-shapes. The calculations in these tables for available
tensile rupture strength assume an effective area, Ae, of 0.75Ag. If the actual effective area is greater than 0.75Ag,
the tabulated values will be conservative and calculations can be performed to obtain higher available strengths. If
the actual effective area is less than 0.75Ag, the tabulated values will be unconservative and calculations are
necessary to determine the available strength.
D3. EFFECTIVE NET AREA
The gross area, Ag, is the total cross-sectional area of the member.
In computing net area, An, AISC Specification Section B4.3 requires that an extra z in. be added to the bolt hole
diameter.
A computation of the effective area for a chain of holes is presented in Example D.9.
Unless all elements of the cross section are connected, Ae = AnU, where U is a reduction factor to account for
shear lag. The appropriate values of U can be obtained from Table D3.1 of the AISC Specification.
D4. BUILT-UP MEMBERS
The limitations for connections of built-up members are discussed in Section D4 of the AISC Specification.
D5. PIN-CONNECTED MEMBERS
An example of a pin-connected member is given in Example D.7.
D6. EYEBARS
An example of an eyebar is given in Example D.8. The strength of an eyebar meeting the dimensional
requirements of AISC Specification Section D6 is governed by tensile yielding of the body.

D-2
EXAMPLE D.1 W-SHAPE TENSION MEMBER
Given:
Select an 8-in. W-shape, ASTM A992, to carry a dead load of 30 kips and a live load of 90 kips in tension. The
member is 25 ft long. Verify the member strength by both LRFD and ASD with the bolted end connection shown.
Verify that the member satisfies the recommended slenderness limit. Assume that connection limit states do not
govern.
Solution:
From Chapter 2 of ASCE/SEI 7, the required tensile strength is:
LRFD ASD
Pu = 1.2(30 kips) + 1.6(90 kips)
= 180 kips
Pa = 30 kips + 90 kips
= 120 kips
From AISC Manual Table 5-1, try a W8×21.
From AISC Manual Table 2-4, the material properties are as follows:
W8×21
ASTM A992
Fy = 50 ksi
Fu = 65 ksi
From AISC Manual Tables 1-1 and 1-8, the geometric properties are as follows:
W8×21
Ag = 6.16 in.2
bf = 5.27 in.
tf = 0.400 in.
d = 8.28 in.
ry = 1.26 in.
WT4×10.5
y = 0.831 in.
Tensile Yielding
From AISC Manual Table 5-1, the tensile yielding strength is:
LRFD ASD
277 kips > 180 kips o.k. 184 kips > 120 kips o.k.

D-3
Tensile Rupture
Verify the table assumption that Ae/Ag ≥ 0.75 for this connection.
Calculate the shear lag factor, U, as the larger of the values from AISC Specification Section D3, Table D3.1 case
2 and case 7.
From AISC Specification Section D3, for open cross sections, U need not be less than the ratio of the gross area of
the connected element(s) to the member gross area.
U =
2 f f
g
b t
A
= 2
2(5.27 in.)(0.400 in.)
6.16 in.
= 0.684
Case 2: Check as two WT-shapes per AISC Specification Commentary Figure C-D3.1, with 0.831 in.x y= =
U =1
x
l
−
=
0.831 in.
1
9.00 in.
−
= 0.908
Case 7:
bf = 5.27 in.
d = 8.28 in.
bf < qd
U = 0.85
Use U = 0.908.
Calculate An using AISC Specification Section B4.3.
An = Ag – 4(dh + z in.)tf
= 6.16 in.2
– 4(m in. + z in.)(0.400 in.)
= 4.76 in.2
Calculate Ae using AISC Specification Section D3.
Ae = AnU (Spec. Eq. D3-1)
= 4.76 in.2
(0.908)
= 4.32 in.2
2
2
4.32 in.
6.16 in.
e
g
A
A
=
= 0.701 < 0.75; therefore, table values for rupture are not valid.
The available tensile rupture strength is,

D-4
Pn = FuAe (Spec. Eq. D2-2)
= 65 ksi(4.32 in.2
)
= 281 kips
From AISC Specification Section D2, the available tensile rupture strength is:
LRFD ASD
φt = 0.75
φtPn = 0.75(281 kips)
= 211 kips
211 kips > 180 kips o.k.
Ωt = 2.00
281 kips
2.00
n
t
P
=
Ω
= 141 kips
Check Recommended Slenderness Limit
25.0 ft 12.0 in.
1.26 in. ft
L
r
⎛ ⎞⎛ ⎞
= ⎜ ⎟⎜ ⎟
⎝ ⎠⎝ ⎠
= 238 < 300 from AISC Specification Section D1 o.k.
The W8×21 available tensile strength is governed by the tensile rupture limit state at the end connection.
See Chapter J for illustrations of connection limit state checks.

D-5
EXAMPLE D.2 SINGLE ANGLE TENSION MEMBER
Given:
Verify, by both ASD and LRFD, the tensile strength of an L4×4×2, ASTM A36, with one line of (4) w-in.-
diameter bolts in standard holes. The member carries a dead load of 20 kips and a live load of 60 kips in tension.
Calculate at what length this tension member would cease to satisfy the recommended slenderness limit. Assume
that connection limit states do not govern.
Solution:
L4×4×2
ASTM A36
Fy = 36 ksi
Fu = 58 ksi
From AISC Manual Table 1-7, the geometric properties are as follows:
L4×4×2
Ag = 3.75 in.2
rz = 0.776 in.
y = 1.18 in. = x
LRFD ASD
Pu = 1.2(20 kips) + 1.6(60 kips)
= 120 kips
= 80.0 kips
Tensile Yielding
Pn = FyAg (Spec. Eq. D2-1)
= 36 ksi(3.75 in.2
)
= 135 kips
From AISC Specification Section D2, the available tensile yielding strength is:
LRFD ASD
φt = 0.90
φtPn = 0.90(135 kips)
= 122 kips
Ωt = 1.67
135 kips
1.67
n
t
P
=
Ω
= 80.8 kips

D-6
Tensile Rupture
Calculate U as the larger of the values from AISC Specification Section D3, Table D3.1 Case 2 and Case 8.
From AISC Specification Section D3, for open cross sections, U need not be less than the ratio of the gross area of
the connected element(s) to the member gross area, therefore,
U = 0.500
Case 2:
U =1
x
l
−
=
1.18in.
1
9.00in.
−
= 0.869
Case 8, with 4 or more fasteners per line in the direction of loading:
U = 0.80
Use U = 0.869.
An = Ag – (dh + z)t
= 3.75 in.2
– (m in. + z in.)(2 in.)
= 3.31 in.2
= 3.31 in.2
(0.869)
= 2.88 in.2
= 58 ksi(2.88 in.2
)
= 167 kips
LRFD ASD
φt = 0.75
φtPn = 0.75(167 kips)
= 125 kips
Ωt = 2.00
167 kips
2.00
n
t
P
=
Ω
= 83.5 kips
The L4×4×2 available tensile strength is governed by the tensile yielding limit state.
LRFD ASD
φtPn = 122 kips
80.8 kipsn
t
P
=
Ω
80.8 kips > 80.0 kips o.k.

D-7
Recommended Lmax
Using AISC Specification Section D1:
Lmax = 300rz
=
ft
(300)(0.776in.)
12.0 in.
⎛ ⎞
⎜ ⎟
⎝ ⎠
= 19.4 ft
Note: The L/r limit is a recommendation, not a requirement.

D-8
EXAMPLE D.3 WT-SHAPE TENSION MEMBER
Given:
A WT6×20, ASTM A992 member has a length of 30 ft and carries a dead load of 40 kips and a live load of 120
kips in tension. The end connection is fillet welded on each side for 16 in. Verify the member tensile strength by
both LRFD and ASD. Assume that the gusset plate and the weld are satisfactory.
Solution:
WT6×20
ASTM A992
Fy = 50 ksi
Fu = 65 ksi
WT6×20
Ag = 5.84 in.2
bf = 8.01 in.
tf = 0.515 in.
rx = 1.57 in.
y = 1.09 in. = x (in equation for U)
LRFD ASD
Pu = 1.2(40 kips) + 1.6(120 kips)
= 240 kips
= 160 kips
Tensile Yielding
Check tensile yielding limit state using AISC Manual Table 5-3.
LRFD ASD
φtPn = 263 kips > 240 kips o.k. n
t
P
Ω
= 175 kips > 160 kips o.k.

D-9
Tensile Rupture
Check tensile rupture limit state using AISC Manual Table 5-3.
LRFD ASD
φtPn = 214 kips < 240 kips n.g. n
t
P
Ω
= 142 kips < 160 kips n.g.
The tabulated available rupture strengths may be conservative for this case; therefore, calculate the exact solution.
Calculate U as the larger of the values from AISC Specification Section D3 and Table D3.1 case 2.
From AISC Specification Section D3, for open cross-sections, U need not be less than the ratio of the gross area
of the connected element(s) to the member gross area.
U =
f f
g
b t
A
= 2
8.01 in.(0.515 in.)
5.84 in.
= 0.706
Case 2:
U =1
x
l
−
=
1.09in.
1
16.0in.
−
= 0.932
Use U = 0.932.
An = Ag (because there are no reductions due to holes or notches)
= 5.84 in.2
= 5.84 in.2
(0.932)
= 5.44 in.2
Calculate Pn.
= 65 ksi(5.44 in.2
)
= 354 kips

D-10
LRFD ASD
φt = 0.75
φtPn = 0.75(354 kips)
= 266 kips
Ωt = 2.00
354 kips
2.00
n
t
P
=
Ω
= 177 kips
Alternately, the available tensile rupture strengths can be determined by modifying the tabulated values. The
available tensile rupture strengths published in the tension member selection tables are based on the assumption
that Ae = 0.75Ag. The actual available strengths can be determined by adjusting the values from AISC Manual
Table 5-3 as follows:
LRFD ASD
φtPn = 214 kips
0.75
e
g
A
A
⎛ ⎞
⎜ ⎟
⎝ ⎠
= 214 kips
( )
2
2
5.44 in.
0.75 5.84 in.
⎛ ⎞
⎜ ⎟
⎜ ⎟
⎝ ⎠
= 266 kips
n
t
P
Ω
= 142 kips
0.75
e
g
A
A
⎛ ⎞
⎜ ⎟
⎝ ⎠
= 142 kips
( )
2
2
5.44 in.
0.75 5.84 in.
⎛ ⎞
⎜ ⎟
⎜ ⎟
⎝ ⎠
= 176 kips
The WT6×20 available tensile strength is governed by the tensile yielding limit state.
LRFD ASD
φtPn = 263 kips
n
t
P
Ω
= 175 kips
Recommended Slenderness Limit
30.0 ft 12.0 in.
1.57 in. ft
L
r
⎛ ⎞⎛ ⎞
= ⎜ ⎟⎜ ⎟
⎝ ⎠⎝ ⎠

D-11
EXAMPLE D.4 RECTANGULAR HSS TENSION MEMBER
Given:
Verify the tensile strength of an HSS6×4×a, ASTM A500 Grade B, with a length of 30 ft. The member is
carrying a dead load of 35 kips and a live load of 105 kips in tension. The end connection is a fillet welded 2-in.-
thick single concentric gusset plate with a weld length of 16 in. Assume that the gusset plate and weld are
satisfactory.
Solution:
ASTM A500 Grade B
Fy = 46 ksi
Fu = 58 ksi
HSS6×4×a
Ag = 6.18 in.2
ry = 1.55 in.
t = 0.349 in.
LRFD ASD
Pu = 1.2(35 kips) + 1.6(105 kips)
= 210 kips
= 140 kips
Tensile Yielding
LRFD ASD
t
P
Ω
= 170 kips > 140 kips o.k.
Tensile Rupture

D-12
LRFD ASD
φtPn = 201 kips < 210 kips n.g. n
t
P
Ω
= 134 kips < 140 kips n.g.
The tabulated available rupture strengths may be conservative in this case; therefore, calculate the exact solution.
Calculate U from AISC Specification Table D3.1 case 6.
x =
2
2
4( )
B BH
B H
+
+
=
( ) ( )( )
( )
2
4.00 in. 2 4.00 in. 6.00 in.
4 4.00 in. 6.00 in.
+
+
= 1.60 in.
U =1
x
l
−
=
1.60in.
1
16.0in.
−
= 0.900
Allowing for a z-in. gap in fit-up between the HSS and the gusset plate:
An = Ag – 2(tp + z in.)t
= 6.18 in.2
– 2(2 in. + z in.)(0.349 in.)
= 5.79 in.2
Ae =AnU (Spec. Eq. D3-1)
= 5.79 in.2
(0.900)
= 5.21 in.2
Calculate Pn.
= 58 ksi(5.21 in2
)
= 302 kips
LRFD ASD
φt = 0.75
φtPn = 0.75(302 kips)
= 227 kips
Ωt = 2.00
302 kips
2.00
n
t
P
=
Ω
= 151 kips
The HSS available tensile strength is governed by the tensile rupture limit state.

D-13
30.0 ft 12.0 in.
1.55 in. ft
L
r
⎛ ⎞⎛ ⎞
= ⎜ ⎟⎜ ⎟
⎝ ⎠⎝ ⎠

D-14
EXAMPLE D.5 ROUND HSS TENSION MEMBER
Given:
Verify the tensile strength of an HSS6×0.500, ASTM A500 Grade B, with a length of 30 ft. The member carries a
dead load of 40 kips and a live load of 120 kips in tension. Assume the end connection is a fillet welded 2-in.-
thick single concentric gusset plate with a weld length of 16 in. Assume that the gusset plate and weld are
satisfactory.
Solution:
ASTM A500 Grade B
Fy = 42 ksi
Fu = 58 ksi
HSS6×0.500
Ag = 8.09 in.2
r = 1.96 in.
t = 0.465 in.
LRFD ASD
Pu = 1.2(40 kips) + 1.6(120 kips)
= 240 kips
= 160 kips
Tensile Yielding
LRFD ASD
t
P
Ω
= 203 kips > 160 kips o.k.
Tensile Rupture

D-15
LRFD ASD
t
P
Ω
= 176 kips > 160 kips o.k.
Check that Ae/Ag ≥ 0.75 as assumed in table.
Determine U from AISC Specification Table D3.1 Case 5.
L = 16.0 in.
D = 6.00 in.
16.0 in.
6.00 in.
L
D
=
= 2.67 > 1.3, therefore U = 1.0
Allowing for a z-in. gap in fit-up between the HSS and the gusset plate,
An = Ag – 2(tp + z in.)t
= 8.09 in.2
– 2(0.500 in. + z in.)(0.465 in.)
= 7.57 in.2
= 7.57 in.2
(1.0)
= 7.57 in.2
2
2
7.57 in.
8.09 in.
e
g
A
A
=
= 0.936 > 0.75 o.k., but conservative
Calculate Pn.
= (58 ksi)(7.57 in.2
)
= 439 kips
LRFD ASD
φt = 0.75
φtPn = 0.75(439 kips)
= 329 kips
Ωt = 2.00
439 kips
2.00
n
t
P
=
Ω
= 220 kips
30.0 ft 12.0 in.
1.96 in. ft
L
r
⎛ ⎞⎛ ⎞
= ⎜ ⎟⎜ ⎟
⎝ ⎠⎝ ⎠

D-16

D-17
EXAMPLE D.6 DOUBLE ANGLE TENSION MEMBER
Given:
A 2L4×4×2 (a-in. separation), ASTM A36, has one line of (8) ¾-in.-diameter bolts in standard holes and is 25 ft
in length. The double angle is carrying a dead load of 40 kips and a live load of 120 kips in tension. Verify the
member tensile strength. Assume that the gusset plate and bolts are satisfactory.
Solution:
ASTM A36
Fy = 36 ksi
Fu = 58 ksi
From AISC Manual Tables 1-7 and 1-15, the geometric properties are as follows:
L4×4×2
Ag = 3.75 in.2
x = 1.18 in.
2L4×4×2 (s = a in.)
ry = 1.83 in.
rx = 1.21 in.
LRFD ASD
Pn = 1.2(40 kips) + 1.6(120 kips)
= 240 kips
Pn = 40 kips + 120 kips
= 160 kips
Tensile Yielding
= 36 ksi(2)(3.75 in.2
)
= 270 kips

D-18
LRFD ASD
φt = 0.90
φtPn = 0.90(270 kips)
= 243 kips
Ωt = 1.67
270 kips
1.67
n
t
P
=
Ω
= 162 kips
Tensile Rupture
Calculate U as the larger of the values from AISC Specification Section D3, Table D3.1 case 2 and case 8.
From AISC Specification Section D3, for open cross-sections, U need not be less than the ratio of the gross area
of the connected element(s) to the member gross area.
U = 0.500
Case 2:
U = 1
x
l
−
=
1.18in.
1
21.0in.
−
= 0.944
Case 8, with 4 or more fasteners per line in the direction of loading:
U = 0.80
Use U = 0.944.
An = Ag – 2(dh + z in.)t
= 2(3.75 in.2
) – 2(m in. + z in.)(2 in.)
= 6.63 in.2
= 6.63 in.2
(0.944)
= 6.26 in.2
Calculate Pn.
= 58 ksi(6.26 in.2
)
= 363 kips

D-19
LRFD ASD
φt = 0.75
φtPn = 0.75(363 kips)
= 272 kips
Ωt = 2.00
363 kips
2.00
n
t
P
=
Ω
= 182 kips
The double angle available tensile strength is governed by the tensile yielding limit state.
LRFD ASD
243 kips > 240 kips o.k. 162 kips > 160 kips o.k.
25.0 ft 12.0 in.
1.21 in. ftx
L
r
⎛ ⎞⎛ ⎞
= ⎜ ⎟⎜ ⎟
⎝ ⎠⎝ ⎠
Note: From AISC Specification Section D4, the longitudinal spacing of connectors between components of built-
up members should preferably limit the slenderness ratio in any component between the connectors to a maximum
of 300.

D-20
EXAMPLE D.7 PIN-CONNECTED TENSION MEMBER
Given:
An ASTM A36 pin-connected tension member with the dimensions shown as follows carries a dead load of 4 kips
and a live load of 12 kips in tension. The diameter of the pin is 1 inch, in a Q-in. oversized hole. Assume that the
pin itself is adequate. Verify the member tensile strength.
Solution:
Plate
ASTM A36
Fy = 36 ksi
Fu = 58 ksi
The geometric properties are as follows:
w = 4.25 in.
t = 0.500 in.
d = 1.00 in.
a = 2.25 in.
c = 2.50 in.
dh = 1.03 in.
Check dimensional requirements using AISC Specification Section D5.2.
1. be = 2t + 0.63 in.
= 2(0.500 in.) + 0.63 in.
= 1.63 in. ≤ 1.61 in.
be = 1.61 in. controls
2. a > 1.33be
2.25 in. > 1.33(1.61 in.)
= 2.14 in. o.k.

D-21
3. w > 2be + d
4.25 in. > 2(1.61 in.) + 1.00 in.
= 4.22 in. o.k.
4. c > a
2.50 in. > 2.25 in. o.k.
LRFD ASD
Pu = 1.2(4 kips) + 1.6(12 kips)
= 24.0 kips
= 16.0 kips
Tensile Rupture
Calculate the available tensile rupture strength on the effective net area.
Pn = Fu(2tbe) (Spec. Eq. D5-1)
= 58 ksi (2)(0.500 in.)(1.61 in.)
= 93.4 kips
From AISC Specification Section D5.1, the available tensile rupture strength is:
LRFD ASD
φt = 0.75
φtPn = 0.75(93.4 kips)
= 70.1 kips
Ωt = 2.00
93.4 kips
2.00
n
t
P
=
Ω
= 46.7 kips
Shear Rupture
Asf = 2t(a + d/2)
= 2(0.500 in.)[2.25 in. + (1.00 in./2)]
= 2.75 in.2
Pn = 0.6FuAsf (Spec. Eq. D5-2)
= 0.6(58 ksi)(2.75 in.2
)
= 95.7 kips
From AISC Specification Section D5.1, the available shear rupture strength is:
LRFD ASD
φsf = 0.75
φsfPn = 0.75(95.7 kips)
= 71.8 kips
Ωsf = 2.00
95.7 kips
2.00
n
sf
P
=
Ω
= 47.9 kips
Bearing
Apb = 0.500 in.(1.00 in.)
= 0.500 in.2

D-22
Rn = 1.8FyApb (Spec. Eq. J7-1)
= 1.8(36 ksi)(0.500 in.2
)
= 32.4 kips
From AISC Specification Section J7, the available bearing strength is:
LRFD ASD
φ = 0.75
φPn = 0.75(32.4 kips)
= 24.3 kips
Ω = 2.00
32.4 kips
2.00
nP
=
Ω
= 16.2 kips
Tensile Yielding
Ag = wt
= 4.25 in. (0.500 in.)
= 2.13 in.2
= 36 ksi (2.13 in.2
)
= 76.7 kips
LRFD ASD
φt = 0.90
φtPn = 0.90(76.7 kips)
= 69.0 kips
Ωt = 1.67
76.7 kips
1.67
n
t
P
=
Ω
= 45.9 kips
The available tensile strength is governed by the bearing strength limit state.
LRFD ASD
φPn = 24.3 kips
24.3 kips > 24.0 kips o.k.
nP
Ω
= 16.2 kips
16.2 kips > 16.0 kips o.k.
See Example J.6 for an illustration of the limit state calculations for a pin in a drilled hole.

D-23
EXAMPLE D.8 EYEBAR TENSION MEMBER
Given:
A s-in.-thick, ASTM A36 eyebar member as shown, carries a dead load of 25 kips and a live load of 15 kips in
tension. The pin diameter, d, is 3 in. Verify the member tensile strength.
Solution:
Plate
ASTM A36
Fy = 36 ksi
Fu = 58 ksi
w = 3.00 in.
b = 2.23 in.
t = s in.
dhead = 7.50 in.
d = 3.00 in.
dh = 3.03 in.
R = 8.00 in.
Check dimensional requirements using AISC Specification Section D6.1 and D6.2.
1. t > 2 in.
s in. > 2 in. o.k.
2. w < 8t
3.00 in. < 8(s in.)
= 5.00 in. o.k.

D-24
3. d > d w
3.00 in. > d(3.00 in.)
= 2.63 in. o.k.
4. dh < d + Q in.
3.03 in. < 3.00 in. + (Q in.)
= 3.03 in. o.k.
5. R > dhead
8.00 in. > 7.50 in. o.k.
6. q w < b < w w
q(3.00 in.) < 2.23 in. < w(3.00 in.)
2.00 in. < 2.23 in. < 2.25 in. o.k.
LRFD ASD
Pu = 1.2(25.0 kips) + 1.6(15.0 kips)
= 54.0 kips
Pa = 25.0 kips + 15.0 kips
= 40.0 kips
Tensile Yielding
Calculate the available tensile yielding strength at the eyebar body (at w).
Ag = wt
= 3.00 in.(s in.)
= 1.88 in.2
= 36 ksi(1.88 in.2
)
= 67.7 kips
LRFD ASD
φt = 0.90
φtPn = 0.90(67.7 kips)
= 60.9 kips
60.9 kips > 54.0 kips o.k.
Ωt = 1.67
67.7 kips
1.67
n
t
P
=
Ω
= 40.5 kips
40.5 kips > 40.0 kips o.k.
The eyebar tension member available strength is governed by the tensile yielding limit state.
Note: The eyebar detailing limitations ensure that the tensile yielding limit state at the eyebar body will control
the strength of the eyebar itself. The pin should also be checked for shear yielding, and, if the material strength is
less than that of the eyebar, bearing.
See Example J.6 for an illustration of the limit state calculations for a pin in a drilled hole.

D-25
EXAMPLE D.9 PLATE WITH STAGGERED BOLTS
Given:
Compute An and Ae for a 14-in.-wide and 2-in.-thick plate subject to tensile loading with staggered holes as
shown.
Solution:
Calculate net hole diameter using AISC Specification Section B4.3.
dnet = dh + z in.
= m in. + z in.
= 0.875 in.
Compute the net width for all possible paths across the plate. Because of symmetry, many of the net widths are
identical and need not be calculated.
2
14.0
4
net
s
w d
g
= − Σ + Σ from AISC Specification Section B4.3.
Line A-B-E-F: w = 14.0 in. − 2(0.875 in.)
= 12.3 in.
Line A-B-C-D-E-F: ( )
( )
( )
( )
( )
2 2
2.50in. 2.50in.
14.0in. 4 0.875in.
4 3.00in. 4 3.00in.
w = − + +
= 11.5 in. controls
Line A-B-C-D-G: ( )
( )
( )
2
2.50in.
14.0in. 3 0.875in.
4 3.00in.
w = − +
= 11.9 in.
Line A-B-D-E-F: ( )
( )
( )
( )
( )
2 2
2.50in. 2.50in.
14.0in. 3 0.875in.
4 7.00in. 4 3.00in.
w = − + +
= 12.1 in.
Therefore, 11.5 in.(0.500 in.)nA =

D-26
= 5.75 in.2
Calculate U.
From AISC Specification Table D3.1 case 1, because tension load is transmitted to all elements by the fasteners,
U = 1.0
= 5.75 in.2
(1.0)
= 5.75 in.2

E-
1
Chapter E
Design of Members for Compression
This chapter covers the design of compression members, the most common of which are columns. The AISC
Manual includes design tables for the following compression member types in their most commonly available
grades.
• wide-flange column shapes
• HSS
• double angles
• single angles
LRFD and ASD information is presented side-by-side for quick selection, design or verification. All of the tables
account for the reduced strength of sections with slender elements.
The design and selection method for both LRFD and ASD designs is similar to that of previous AISC
Specifications, and will provide similar designs. In this AISC Specification, ASD and LRFD will provide identical
designs when the live load is approximately three times the dead load.
The design of built-up shapes with slender elements can be tedious and time consuming, and it is recommended
that standard rolled shapes be used, when possible.
E1. GENERAL PROVISIONS
The design compressive strength, φcPn, and the allowable compressive strength, Pn/Ωc, are determined as follows:
Pn = nominal compressive strength based on the controlling buckling mode
φc = 0.90 (LRFD) Ωc = 1.67 (ASD)
Because Fcr is used extensively in calculations for compression members, it has been tabulated in AISC Manual
Table 4-22 for all of the common steel yield strengths.
E2. EFFECTIVE LENGTH
In the AISC Specification, there is no limit on slenderness, KL/r. Per the AISC Specification Commentary, it is
recommended that KL/r not exceed 200, as a practical limit based on professional judgment and construction
economics.
Although there is no restriction on the unbraced length of columns, the tables of the AISC Manual are stopped at
common or practical lengths for ordinary usage. For example, a double L3×3×¼, with a a-in. separation has an ry
of 1.38 in. At a KL/r of 200, this strut would be 23’-0” long. This is thought to be a reasonable limit based on
fabrication and handling requirements.
Throughout the AISC Manual, shapes that contain slender elements when supplied in their most common material
grade are footnoted with the letter “c”. For example, see a W14×22c
.
E3. FLEXURAL BUCKLING OF MEMBERS WITHOUT SLENDER ELEMENTS

E-
2
Nonslender sections, including nonslender built-up I-shaped columns and nonslender HSS columns, are governed
by these provisions. The general design curve for critical stress versus KL/r is shown in Figure E-1.
The term L is used throughout this chapter to describe the length between points that are braced against lateral
and/or rotational displacement.
TRANSITION POINT LIMITING VALUES OF KL/r
yF , ksi Limiting KL/r 0.44 yF , ksi
36 134 15.8
50 113 22.0
60 104 26.4
70 96 30.8
E4. TORSIONAL AND FLEXURAL-TORSIONAL BUCKLING OF MEMBERS WITHOUT SLENDER
ELEMENTS
This section is most commonly applicable to double angles and WT sections, which are singly-symmetric shapes
subject to torsional and flexural-torsional buckling. The available strengths in axial compression of these shapes
are tabulated in Part 4 of the AISC Manual and examples on the use of these tables have been included in this
chapter for the shapes with KLz = KLy.
E5. SINGLE ANGLE COMPRESSION MEMBERS
The available strength of single angle compression members is tabulated in Part 4 of the AISC Manual.
E6. BUILT-UP MEMBERS
There are no tables for built-up shapes in the AISC Manual, due to the number of possible geometries. This
section suggests the selection of built-up members without slender elements, thereby making the analysis
relatively straightforward.
Fig. E-1. Standard column curve.
E3-3
Elastic
buckling
E3-2
Inelastic
buckling
Transition between
equations (location
varies by Fy)
Critical
stress,
ksi
KL/r

E-
3
E7. MEMBERS WITH SLENDER ELEMENTS
The design of these members is similar to members without slender elements except that the formulas are
modified by a reduction factor for slender elements, Q. Note the similarity of Equation E7-2 with Equation E3-2,
and the similarity of Equation E7-3 with Equation E3-3.
The tables of Part 4 of the AISC Manual incorporate the appropriate reductions in available strength to account
for slender elements.
Design examples have been included in this Chapter for built-up I-shaped members with slender webs and slender
flanges. Examples have also been included for a double angle, WT and an HSS shape with slender elements.

E-
4
EXAMPLE E.1A W-SHAPE COLUMN DESIGN WITH PINNED ENDS
Given:
Select an ASTM A992 (Fy = 50 ksi) W-shape column to carry an axial dead load of
140 kips and live load of 420 kips. The column is 30 ft long and is pinned top and
bottom in both axes. Limit the column size to a nominal 14-in. shape.
Solution:
From Chapter 2 of ASCE/SEI 7, the required compressive strength is:
LRFD ASD
Pu = 1.2(140 kips) + 1.6(420 kips)
= 840 kips
= 560 kips
Column Selection
From AISC Specification Commentary Table C-A-7.1, for a pinned-pinned condition, K = 1.0.
Because the unbraced length is the same in both the x-x and y-y directions and rx exceeds ry for all W-shapes, y-y
axis bucking will govern.
Enter the table with an effective length, KLy, of 30 ft, and proceed across the table until reaching the least weight
shape with an available strength that equals or exceeds the required strength. Select a W14×132.

E-
5
From AISC Manual Table 4-1, the available strength for a y-y axis effective length of 30 ft is:
LRFD ASD
φcPn = 893 kips > 840 kips o.k. n
c
P
Ω
= 594 kips > 560 kips o.k.

E-
6
EXAMPLE E.1B W-SHAPE COLUMN DESIGN WITH INTERMEDIATE BRACING
Given:
Redesign the column from Example E.1A assuming the column is
laterally braced about the y-y axis and torsionally braced at the midpoint.
Solution:
LRFD ASD
Pu = 1.2(140 kips) + 1.6(420 kips)
= 840 kips
= 560 kips
Column Selection
Because the unbraced lengths differ in the two axes, select the member using the y-y axis then verify the strength
in the x-x axis.
Enter AISC Manual Table 4-1 with a y-y axis effective length, KLy, of 15 ft and proceed across the table until
reaching a shape with an available strength that equals or exceeds the required strength. Try a W14×90. A 15 ft
long W14×90 provides an available strength in the y-y direction of:
LRFD ASD
φcPn = 1,000 kips n
c
P
Ω
= 667 kips
The rx /ry ratio for this column, shown at the bottom of AISC Manual Table 4-1, is 1.66. The equivalent y-y axis
effective length for strong axis buckling is computed as:
30.0ft
1.66
KL =
= 18.1 ft

E-
7
From AISC Manual Table 4-1, the available strength of a W14×90 with an effective length of 18 ft is:
LRFD ASD
φcPn = 929 kips > 840 kips o.k. n
c
P
Ω
= 618 kips > 560 kips o.k.
The available compressive strength is governed by the x-x axis flexural buckling limit state.
The available strengths of the columns described in Examples E.1A and E.1B are easily selected directly from the
AISC Manual Tables. The available strengths can also be verified by hand calculations, as shown in the following
Examples E.1C and E.1D.

E-
8
EXAMPLE E.1C W-SHAPE AVAILABLE STRENGTH CALCULATION
Given:
Calculate the available strength of a W14×132 column with unbraced lengths of 30 ft in both axes. The material
properties and loads are as given in Example E.1A.
Solution:
ASTM A992
Fy = 50 ksi
Fu = 65 ksi
W14×132
Ag = 38.8 in.2
rx = 6.28 in.
ry = 3.76 in.
Slenderness Check
Because the unbraced length is the same for both axes, the y-y axis will govern.
( )1.0 30.0 ft 12.0in
3.76 in. ft
y y
y
K L
r
⎛ ⎞⎛ ⎞
= ⎜ ⎟⎜ ⎟
⎝ ⎠⎝ ⎠
= 95.7
For Fy = 50 ksi, the available critical stresses, φcFcr and Fcr/Ωc for KL/r = 95.7 are interpolated from AISC Manual
Table 4-22 as follows:
LRFD ASD
φcFcr = 23.0 ksi
( )2
38.8 in. 23.0 ksic nPφ =
= 892 kips > 840 kips o.k.
cr
c
F
Ω
= 15.4 ksi
( )2
38.8 in. 15.4 ksin
c
P
=
Ω
= 598 kips > 560 kips o.k.
Note that the calculated values are approximately equal to the tabulated values.

E-
9
EXAMPLE E.1D W-SHAPE AVAILABLE STRENGTH CALCULATION
Given:
Calculate the available strength of a W14×90 with a strong axis unbraced length of 30.0 ft and weak axis and
torsional unbraced lengths of 15.0 ft. The material properties and loads are as given in Example E.1A.
Solution:
ASTM A992
Fy = 50 ksi
Fu = 65 ksi
W14×90
Ag = 26.5 in.2
rx = 6.14 in.
ry = 3.70 in.
Slenderness Check
( )1.0 30.0 ft 12in.
6.14 in. ft
x
x
KL
r
⎛ ⎞
= ⎜ ⎟
⎝ ⎠
= 58.6 governs
( )1.0 15.0 ft 12 in.
3.70 in. ft
y
y
KL
r
⎛ ⎞
= ⎜ ⎟
⎝ ⎠
= 48.6
Critical Stresses
The available critical stresses may be interpolated from AISC Manual Table 4-22 or calculated directly as
follows:
Calculate the elastic critical buckling stress, Fe.
2
2e
E
F
KL
r
π
=
⎛ ⎞
⎜ ⎟
⎝ ⎠
(Spec. Eq. E3-4)
( )
( )
2
2
29,000ksi
58.6
π
=
= 83.3 ksi
Calculate the flexural buckling stress, Fcr.

E-
10
29,000ksi
4.71 4.71
50ksiy
E
F
=
= 113
Because 58.6 113,
KL
r
= ≤
0.658
y
e
F
F
cr yF F
⎡ ⎤
= ⎢ ⎥
⎢ ⎥⎣ ⎦
(Spec. Eq. E3-2)
50.0 ksi
83.3 ksi
0.658 50.0ksi
⎡ ⎤
= ⎢ ⎥
⎣ ⎦
= 38.9 ksi
Nominal Compressive Strength
Pn = FcrAg (Spec. Eq. E3-1)
= ( )2
38.9ksi 26.5in.
= 1,030 kips
From AISC Specification Section E1, the available compressive strength is:
LRFD ASD
φc = 0.90
φcPn = 0.90(1,030 kips)
= 927 kips > 840 kips o.k.
Ωc = 1.67
1,030 kips
1.67
n
c
P
=
Ω
= 617 kips > 560 kips o.k.

E-
11
EXAMPLE E.2 BUILT-UP COLUMN WITH A SLENDER WEB
Given:
Verify that a built-up, ASTM A572 Grade 50 column with PL1 in. × 8 in. flanges and a PL4 in. × 15 in. web is
sufficient to carry a dead load of 70 kips and live load of 210 kips in axial compression. The column length is 15
ft and the ends are pinned in both axes.
Solution:
Built-Up Column
ASTM A572 Grade 50
Fy = 50 ksi
Fu = 65 ksi
Built-Up Column
d = 17.0 in.
bf = 8.00 in.
tf = 1.00 in.
h = 15.0 in.
tw = 4 in.
LRFD ASD
Pu = 1.2(70 kips) + 1.6(210 kips)
= 420 kips
= 280 kips
Built-Up Section Properties (ignoring fillet welds)
A = 2(8.00 in.)(1.00 in.) + 15.0 in.(4 in.)
= 19.8 in.2

E-
12
( )( ) ( )
3 3
2 1.00 in. 8.00 in. 15.0 in. in.
12 12
yI = +
4
= 85.4 in.4
y
y
I
r
A
=
4
2
85.4 in.
19.8 in.
=
= 2.08 in.
( )( )
( )( ) ( )( )
3
2
3 3
22
4
12
in. 15.00 in. 2 8.00 in. 1.00 in.
2 8.00 in. 8.00 in. + +
12 12
1,100 in.
x
bh
I Ad= +
=
=
∑ ∑
4
Elastic Flexural Buckling Stress
Because the unbraced length is the same for both axes, the y-y axis will govern by inspection.
( )1.0 15.0 ft 12.0in.
2.08 in. ft
y
y
KL
r
⎛ ⎞
= ⎜ ⎟
⎝ ⎠
= 86.5
Fe =
2
2
E
KL
r
π
⎛ ⎞
⎜ ⎟
⎝ ⎠
(Spec. Eq. E3-4)
=
( )
2
2
(29,000 ksi)
86.5
π
= 38.3 ksi
Elastic Critical Torsional Buckling Stress
Note: Torsional buckling generally will not govern if KLy ≥ KLz; however, the check is included here to illustrate
the calculation.
From the User Note in AISC Specification Section E4,
Cw =
2
4
y oI h
=
4 2
85.4 in. (16.0 in.)
4
= 5,470 in.6

E-
13
From AISC Design Guide 9, Equation 3.4,
J =
3
3
bt
∑
=
( )( ) ( )( )
3 3
2 8.00 in. 1.00 in. + 15.0 in. in.
3
4
= 5.41 in4
Fe =
( )
2
2
1
+
+
w
x yz
EC
GJ
I IK L
⎡ ⎤π
⎢ ⎥
⎢ ⎥⎣ ⎦
(Spec. Eq. E4-4)
=
( )( )
( )( )( )
( )( )
2 6
4
2 4 4
29,000ksi 5,470in. 1
+ 11,200ksi 5.41in.
1,100in. 85.4in.1.0 15ft 12
⎡ ⎤π ⎛ ⎞⎢ ⎥⎜ ⎟
+⎢ ⎥⎝ ⎠⎡ ⎤⎣ ⎦⎣ ⎦
= 91.9 ksi > 38.3 ksi
Therefore, the flexural buckling limit state controls.
Use Fe = 38.3 ksi.
Slenderness
Check for slender flanges using AISC Specification Table B4.1a, then determine Qs, the unstiffened element
(flange) reduction factor using AISC Specification Section E7.1.
Calculate kc using AISC Specification Table B4.1b note [a].
kc =
4
wh t
=
4
15.0 in. in.4
= 0.516, which is between 0.35 and 0.76
For the flanges,
b
t
λ =
=
4.00in.
1.00in.
= 4.00
Determine the flange limiting slenderness ratio, λr, from AISC Specification Table B4.1a case 2
0.64
c
r
y
k E
F
λ =
=
( )0.516 29,000 ksi
0.64
50 ksi
= 11.1
rλ < λ ; therefore, the flange is not slender and Qs = 1.0.

E-
14
Check for a slender web, then determine Qa, the stiffened element (web) reduction factor using AISC
Specification Section E7.2.
h
t
λ =
15.0 in.
in.
=
4
= 60.0
Determine the slender web limit from AISC Specification Table B4.1a case 5
1.49r
y
E
F
λ =
29,000 ksi
1.49
50 ksi
35.9
=
=
rλ > λ ; therefore, the web is slender
Qa = e
g
A
A
(Spec. Eq. E7-16)
where Ae = effective area based on the reduced effective width, be
For AISC Specification Equation E7-17, take f as Fcr with Fcr calculated based on Q = 1.0.
Select between AISC Specification Equations E7-2 and E7-3 based on KL/ry.
KL/r = 86.5 as previously calculated
4.71
y
E
QF
=
( )
29,000ksi
4.71
1.0 50ksi
= 113 > 86.5
Because
KL
r
M 4.71 ,
y
E
QF
Fcr 0.658
y
e
QF
F
yQ F
⎡ ⎤
= ⎢ ⎥
⎢ ⎥⎣ ⎦
(Spec. Eq. E7-2)
( )
( )
1.0 50 ksi
38.3 ksi
1.0 0.658 50 ksi
⎡ ⎤
= ⎢ ⎥
⎢ ⎥⎣ ⎦
= 29.0 ksi
( )
0.34
1.92 1 , wheree
E E
b t b b h
f b t f
⎡ ⎤
= − ≤ =⎢ ⎥
⎢ ⎥⎣ ⎦
(Spec. Eq. E7-17)
( )
( )
29,000 ksi 0.34 29,000 ksi
1.92 in. 1 15.0 in.
29.0 ksi 15.0 in. in. 29.0 ksi
⎡ ⎤
= − ≤⎢ ⎥
⎢ ⎥⎣ ⎦
4
4

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